Solve a non-linear ODE of third order

In summary, the conversation discusses attempting to solve the equation y''' + (1-x^2)y=0 with initial conditions y(0)=0, y'(0)=0, y''(0)=0 using WolframAlpha and MATLAB's ODE45. The conclusion is that the equation cannot be solved analytically and the solution obtained numerically may not be reliable due to the locality of differentiation.
  • #1
SemM
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Hi, I tried to solve the following in Wolfram alpha:

y''' + (1-x^2)y=0

y(0)=0
y'(0)=0
y''(0)=0

however, I got answer which cannot be reproduced (even at wolfram pages).

I have tried ODE45 in MATLAB, but it only gives a plot.

Is there any way to solve this analytically or numerically to give a solution?

Thanks
 
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  • #2
I find the plots in the first and last row give a good answer:
http://www.wolframalpha.com/input/?i=y'''+++(1-x^2)y=0
You only have to draw that ##y=1## line at ##y=0## for the first row plot. It doesn't appear to be solvable analytically. It is already a complicated formula without the ##1-x^2## factor. Maybe you can get an expression with Lambert's W-function, but note, that this isn't a function.
 
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  • #3
SemM said:
y''' + (1-x^2)y=0
y(0)=0
y'(0)=0
y''(0)=0
is satisfied for ##y(x) = 0## :smile:

With nonzero initial conditions it looks like a candidate for numerics to me:

upload_2018-2-20_17-52-42.png


[edit] ah, @fresh_42 did the same but faster.

I'd go for a runge kutta and see what it brings, but I suppose ODE45 is equivalent
 

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  • #4
BvU said:
is satisfied for ##y(x) = 0## :smile:

With nonzero initial conditions it looks like a candidate for numerics to me:

View attachment 220733

[edit] ah, @fresh_42 did the same but faster.

I'd go for a runge kutta and see what it brings, but I suppose ODE45 is equivalent
fresh_42 said:
I find the plots in the first and last row give a good answer:
http://www.wolframalpha.com/input/?i=y'''+++(1-x^2)y=0
You only have to draw that ##y=1## line at ##y=0## for the first row plot. It doesn't appear to be solvable analytically. It is already a complicated formula without the ##1-x^2## factor. Maybe you can get an expression with Lambert's W-function, but note, that this isn't a function.
Thanks for answers.

I have tried ODE45:

fun = @(x,y)[y(2);y(3);y(1)*(2+x^2)];
y0 = [0 0 0];
xspan = [0 5];
[X,Y] = ode45(fun,xspan,y0);
plot(X,Y(:,1),X,Y(:,2),X,Y(:,3));

It is a rather boring plot, and it doesn't appear as what I was looking for.

About the Runge-Kutta, can you send me a link to an example ?

@Fresh42, I got that result earlier indeed, however, I got an even nicer results (but only once!):

Z = (1./3).*(exp(x) + 2*exp(-x/2)*cos(w))

where w = (sqrt(3)*x)./2Note the peculiar properties of this function, it is the same as it third derivative, its sixth derivative and its ninth derivative. It cycles back every third derivation. Unusual for such a long form, I expect.

I am however NOT able to reproduce that solution in Wolfram alpha, as it appeared once in the result as a solution, in addition to the plots. Ttrying to test it in the ODE, with IC y(0)=0, y'(0)=0, y''(0)=0 does not work. So I wonder where that Z function came from..

Thanks
 
  • #5
The last plot on the WolframAlpha page has a big advantage: It gives the correct impression on possible solutions. As you've fixed your initial conditions around ##x=0##, the solution is pretty well determined at this point and the further away, the higher are the uncertainties. This demonstrates once more the locality of differentiation. This means in return, that only in a neighborhood of this point you can get a good convergence of the power series.
 
  • #6
fresh_42 said:
The last plot on the WolframAlpha page has a big advantage: It gives the correct impression on possible solutions. As you've fixed your initial conditions around ##x=0##, the solution is pretty well determined at this point and the further away, the higher are the uncertainties. This demonstrates once more the locality of differentiation. This means in return, that only in a neighborhood of this point you can get a good convergence of the power series.
:(

Thanks Fresh!
 

What is a non-linear ODE of third order?

A non-linear ODE of third order is a type of differential equation that involves a third derivative of the dependent variable, and the dependent variable is raised to a power or multiplied by a function of itself.

How do you solve a non-linear ODE of third order?

To solve a non-linear ODE of third order, you can use various methods such as substitution, power series, or numerical methods like Euler's method. The specific method used will depend on the specific equation and its complexity.

What is the difference between a linear and non-linear ODE of third order?

A linear ODE of third order can be written in the form of a polynomial, where the dependent variable and its derivatives have a power of one. A non-linear ODE of third order, on the other hand, cannot be written in this form and may involve products or powers of the dependent variable and its derivatives.

What are some applications of solving non-linear ODEs of third order?

Non-linear ODEs of third order are commonly used in mathematical modeling of real-world systems such as population growth, chemical reactions, and mechanical systems. They are also used in fields like engineering, physics, and biology.

Are there any challenges in solving non-linear ODEs of third order?

Yes, solving non-linear ODEs of third order can be challenging due to their complexity and the lack of general methods for solving them. Finding exact solutions may not always be possible, and numerical approximations may be needed. Additionally, the initial or boundary conditions for the equation must be specified for a specific solution to be obtained.

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